Difference between revisions of "2020 AMC 8 Problems/Problem 7"
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/61c1MR9tne8 | https://youtu.be/61c1MR9tne8 | ||
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+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=251 | ||
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+ | ~Interstigation | ||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=6|num-a=8}} | {{AMC8 box|year=2020|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:29, 18 April 2021
Contents
Problem
How many integers between and have four distinct digits arranged in increasing order? (For example, is one integer.)
Solution 1
Firstly, observe that the second digit of such a number cannot be or , because the digits must be distinct and increasing. The second digit also cannot be as the number must be less than , so it must be . It remains to choose the latter two digits, which must be distinct digits from . That can be done in ways; there is then only way to order the digits, namely in increasing order. This means the answer is .
Solution 2 (without using the "choose" function)
As in Solution 1, we find that the first two digits must be , and the third digit must be at least . If it is , then there are choices for the last digit, namely , , , , or . Similarly, if the third digit is , there are choices for the last digit, namely , , , and ; if , there are choices; if , there are choices; and if , there is choice. It follows that the total number of such integers is .
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=251
~Interstigation
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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